\( \Large x^{2}-8x+15=0; y^{2}-3y+2=0 \)


A) If x>y

B) \( \Large If x\ge y \)

C) lfx

D) \( \Large If x \le y \)

Correct Answer:
A) If x>y

Description for Correct answer:

\( \Large x^{2} - 8x + 15 = 0 \)

=> \( \Large x^{2} - 5x - 3x + 15 = 0 \)

=> \( \Large x^{2} - 5x - 3x + 15 = 0 \)

=> \( \Large x \left(x-5\right) -3 \left(x-5\right) = 0 \)

=> \( \Large \left(x-5\right) \left(x-3\right) = 0 \)

Therefore, x = 5, 3

and \( \Large y^{2} - 3y + 2 = 0 \)

= \( \Large y^{2} - 2y - y + 2 = 0 \)

= \( \Large y \left(y - 2\right) - 1 \left(y - 2\right) = 0 \)

= y = 2, 1

Therefore, x> y


Part of solved Quadratic Equations questions and answers : >> Elementary Mathematics >> Quadratic Equations








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